JEE Mains · Physics · STD 11 - 7. gravitation
A spaceship orbits around a planet at a height of \(20\,km\) from its surface. Assuming that only gravitational field of the plant acts on the spaceship. What will be the number of complete revolutions made by the spaceship in \(24\,hours\) around the plane? [Given: Mass of plane \(= 8 \times 10^{22}\,kg,\) Radius of planet \(= 2\times 10^6\,m,\) Gravitational constant \(G = 6.67\times 10^{-11}\,Mn^2/kg^2\) ]
- A \(9\)
- B \(11\)
- C \(13\)
- D \(17\)
Answer & Solution
Correct Answer
(B) \(11\)
Step-by-step Solution
Detailed explanation
\(\frac{{m{V^2}}}{r} = \frac{{GMm}}{{{r^2}}}\) \(V = \sqrt {\frac{{GM}}{r}} \) \(n = \frac{{VT}}{{2\pi r}} = \sqrt {\frac{{GM}}{r}} \frac{T}{{2\pi r}}\)…
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